Thread: Partial Fraction and integrate

ok it says to integrate. ....

Integral (5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2) dx

so since the numerator's degree is higher we have to do long divison right? i got 5x + 3 + (x^2+4)/(x^3-x^2)

but now... what do we use to integrate? is it just the fraction portion of the answer to long division? im stuck...

Re:

thanks but how did you get a 9?

Originally Posted by runner07

ok it says to integrate. ....

Integral (5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2) dx

so since the numerator's degree is higher we have to do long divison right? i got 5x + 3 + (x^2+4)/(x^3-x^2)

but now... what do we use to integrate? is it just the fraction portion of the answer to long division? im stuck...

In your long division you should have got
5x +3 +[(5x^2 +4) /(x^3 -x^2)

You forgot the coefficient 5 of the x^2 term.

Now you need to decompose the rational or fraction portion of the integrand, because the degree of the numerator now is lower than that of the degree of the denominator. You cannot use long division anymore.

Your title for your question is partial fraction, remember?

(5x^2 +4) /(x^3 -x^2)
(5x^2 +4) / (x^2)(x -1)

(5x^2 +4) / (x^2)(x-1) = A/x +B/(x^2) +C/(x-1)
Multiply both sides by (x^2)(x-1),
5x^2 +4 = A(x)(x-1) +B(x-1) +C(x^2) -------------(i)

When x = 0, in (i),
0 +4 = 0 +B(-1) +0
So, B = -4 ---------------------**

When x = 1, in (i),
5(1^2) +4 = 0 +0 +C(1^2)
So, C = 9 -------------------------**

When x = 2, and B = -4, and C = 9, in (i),
5(2^2) +4 = A(2)(2-1) +(-4)(2-1) +9(2^2)
24 = 2A -4 +36
24 +4 -36 = 2A
A = -8/2 = -4 -------------------**

Therefore,
INT.[(5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2)]dx

= INT.[5x +3 -4/x -4/(x^2) +9/(x-1)]dx

And so on, and so forth.....